Solution: We recall that for the given boundary value problem (BVP), we say $\lambda $ is an eigenvalue of the given BVP if the solution corresponding to the $\lambda $ is non trivial. The solution is called the eigenfunction corresponding to the eigenvalue $\lambda $.

Let $\lambda \lt 0$. We take $\lambda = -k^2$, where $k$ is a real number. Then the solution of the differential equation \[ y^{\prime\prime} - k^2 y = 0 \] is \[ y(x) = Ae^{kx} + B e^{-kx}. \] Now we will use the boundary condition to find the values of $A$ and $B$. Note that \[ y^\prime (x) = Ake^{kx} - Bke^{-kx} . \]We have \begin{align*} y(0) = 0 & \implies A + B = 0\\ y^\prime (\pi) = 0 & \implies Ake^{k\pi} - Bke^{-k\pi} = 0 \\ & \implies A e^{k\pi} - Be^{-k\pi} = 0. \end{align*} This is a system of linear homogeneous equation and finding the determinant of the coefficient matrix will help to find the solution. The determinant of the coefficient matrix will be \[ \det \begin{pmatrix} 1 & 1 \\ e^{k\pi} & -e^{-k\pi} \\ \end{pmatrix} = -e^{-k\pi} - e^{k\pi} = - \left( e^{k\pi} + \frac{1}{e^{k\pi}} \right) \neq 0. \] Since, the determinant is nonzero, the system has trivial solution and hence, $A=B=0$. Therefore, the solution of the differential equation will be trivial, that is $y\equiv 0$. Hence, if $\lambda \lt 0$, it can not be an eigenvalue.

Now take $\lambda =0$. Then the solution of the differential equation \[ y^{\prime\prime} = 0 \] will be \[ y(x) = Ax + B. \] Applying the boundary conditions, we have \begin{align*} y(0) = 0 \implies B = 0 \\ y^\prime (\pi) = 0 \implies A = 0. \end{align*} Hence, the solution is trivial and therefore, $\lambda =0$ is also not an eigenvalue of the given BVP.

Finally, take $\lambda >0$. Let $\lambda =k^2,~k\in \mathbb{R} $. Then the differential equation will be \[ y^{\prime\prime} + k^2 y = 0 \] and the solution will be \[ y(x) = A\cos kx + B \sin kx. \] Note that \[ y^\prime (x) = -Ak \sin kx + Bk\cos kx. \] Applying the boundary conditions, we have \begin{align*} y(0) = 0 & \implies A = 0 \\ y^\prime (\pi) = 0 & \implies Bk \cos (k\pi) = 0 \\ & \implies B \cos(k\pi) = 0. \end{align*} If $B=0$, then again the solution will be trivial and hence we will not get the eigenvalue. Let $\cos k\pi = 0$. This implies $k = \frac{2n+1}{2}$, where $n$ is an integer. Thus, \[ \lambda _n = k^2 = \left( \frac{2n+1}{2} \right) ^2,~n\in \mathbb{Z} . \] Note that the above collection is same as \[ \lambda _n = \left( \frac{2n+1}{2} \right) ^2,~n= 0,1,2,\ldots . \] Therefore, the eigenvalues of the given BVP will be \[ \left\{ \lambda _n = \left(\frac{2n+1}{2} \right) ^2 : n = 0,1,2,\ldots . \right\} \] Note that $\lambda _n$ is an increasing function of $n$ and hence the difference between the least two eigenvalues will be \[ \lambda _1 - \lambda _0 = \frac{9}{4} - \frac{1}{4} = \frac{8}{4} = 2. \]